Let $A$ be an abelian finitely generated free group and $A/B$ be a torsion group. Show that $rank(A)=rank(B)$.

Let $$A$$ be an abelian free group that is finitely generated, and let $$B\subset A$$ be a subgroup of $$A$$ such that $$A/B$$ is a torsion group. Show that $$rank(A)=rank(B)$$.

From the hypothesis, I know that $$A=$$, so $$rank(A)=n$$. Since the set $$\{x_1, ..., x_n\}$$ clearly generates $$B$$, then I have that $$rank(B)\leq n$$. Now, the idea that I have is to show that if $$S\subset B$$ generates $$B$$, then it also generates $$A$$, but I don't know if this is true, and I don't know how to prove it. I was trying to do this, and here's what I got:

Since $$A/B$$ is a torsion group, for $$a\in A$$, there exists a natural number $$m$$ such that $$ma\in B$$, so $$ma$$ can be written as a linear combination of elements in $$S$$. (I don't know if this is useful).

Any hint would be very appreciated! Thank you!

• Use $\langle X\rangle$ for $\langle X\rangle$ and use $\operatorname{rank}(C)$ for $\operatorname{rank}(C)$. – Shaun Dec 11 '18 at 4:04
• The latter is helpful when, for instance, one needs to juxtapose the rank of something by a symbol on the left. – Shaun Dec 11 '18 at 4:06
• Do you know the structure theorem for modules over PID or the smith normal form? – Dante Grevino Dec 11 '18 at 4:28

I think your idea is good. Let $$\{x_1,\ldots,x_n\}$$ be a $$\mathbb{Z}$$-basis of $$A$$. By hypothesis, the quotient $$A/B$$ is a torsion group so for every $$i$$ there exists a natural number $$m_i$$ such that $$m_ix_i$$ is in $$B$$. Now, the set $$\{m_1x_1,\ldots,m_nx_n\}$$ is $$\mathbb{Z}$$-linearly independent and thus it is a $$\mathbb{Z}$$-basis of a submodule $$C\subseteq B$$ of rank $$n$$. By monotonicity of the rank (over the commutative ring $$\mathbb{Z}$$!), it follows that the rank of $$B$$ is $$n$$.
You can show $$B$$ has finite index in $$A$$, say $$[A:B]=n$$. Then the map $$a\rightarrow a^n$$ is a monomorphism from $$A$$ to $$B$$, so rank ($$A$$)$$\le$$rank($$B$$). The reverse inequality is also true and that gives you what you want.